In any ΔABC, prove that

Question:

In any ΔABC, prove that

$a \sin A-b \sin B=c \sin (A-B)$

 

Solution:

Need to prove: $a \sin A-b \sin B=c \sin (A-B)$

Left hand side,

$=a \sin A-b \sin B$

$=(b \cos C+c \cos B) \sin A-(c \cos A+a \cos C) \sin B$

$=b \cos C \sin A+c \cos B \sin A-c \cos A \sin B-a \cos C \sin B$

$=c(\sin A \cos B-\cos A \sin B)+\cos C(b \sin A-a \sin B)$

We know that, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R$ where R is the circumradius.

Therefore,

$=c(\sin A \cos B-\cos A \sin B)+\cos C(2 R \sin B \sin A-2 R \sin A \sin B)$

$=c(\sin \mathrm{A} \cos \mathrm{B}-\cos \mathrm{A} \sin \mathrm{B})$

$=c \sin (A-B)$

$=$ Right hand side. [Proved]

 

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